$$\int_{0}^{2\pi}e^{e^{it}} \,dt$$
I'm trying to calculate the above integral. I've been told as a hint to see it as an integral along a path, so I've tried to think of it as the integral along a circle with centre 0, radius 1. Since it is a closed path, then the integral would be 0?
I'm self teaching complex analysis and it is new to me, so is this correct? Thanks!
Best Answer
You have the right idea, but be careful how you make the substitution. As $t$ goes from $0$ to $2\pi$, $e^{it}$ will go around the unit circle, so you want to substitute $z=e^{it}$. However, you also get $dz=ie^{it}\,dt=iz\,dt$ so $dt$ becomes $\frac{dz}{iz}$ and the whole integral becomes $$\int_C \frac{e^z}{iz}\,dz$$ where $C$ is the unit circle. Because of the $z$ in the denominator, this integrand has a pole at $z=0$ and is not holomorphic on the unit disk, so you cannot conclude that the integral is $0$. Instead, by the Cauchy integral formula, $$\frac{1}{2\pi i}\int_C\frac{e^z}{z}\,dz$$ is just the value of the function $e^z$ at $0$, which is $1$. Comparing this to your integral, you get that your integral is $2\pi$.